Problem: The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered. 

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x, y, r, where (x, y) are the coordinates of the stadium's center ( - 103 ≤ x, y ≤ 103), and r (1 ≤ r ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line. 

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

Examples

Input

0 0 10
60 0 10
30 30 10


Output

30.00000 0.00000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double Maxalpha = -1;
const double PI = acos(-1.);
const double EPS = 1E-10;
int Zero(double O) { return fabs(O) <= EPS; }
void Read(double &now) {
  double IO;
  scanf("%lf", &IO);
  now = IO;
}
struct Point {
  double x, y;
  void read() {
    Read(x);
    Read(y);
  }
  void Print() { printf("%.5f %.5f\n", (double)x, (double)y); }
  void init(double _x, double _y) {
    x = _x;
    y = _y;
  }
  double operator*(const Point &b) const { return x * b.y - y * b.x; }
  Point operator*(const double &b) const { return (Point){x * b, y * b}; }
  double operator^(const Point &b) const { return x * b.x + y * b.y; }
  Point operator-(const Point &b) const { return (Point){x - b.x, y - b.y}; }
  Point operator+(const Point &b) const { return (Point){x + b.x, y + b.y}; }
  int operator<(const Point &b) const { return x < b.x || x == b.x && y < b.y; }
  int operator==(const Point &b) const {
    return Zero(x - b.x) && Zero(y - b.y);
  }
  void Rotate(double alpha) {
    double _x = x * cos(alpha) - y * sin(alpha);
    double _y = y * cos(alpha) + x * sin(alpha);
    x = _x;
    y = _y;
  }
} ans, ERRORPOINT = (Point){1e99, 1e99};
struct Line {
  Point p, q;
  void Print() {
    p.Print();
    q.Print();
    puts("");
  }
};
struct Circle {
  Point O;
  double r;
  void read() {
    O.read();
    Read(r);
  }
  void Print() {
    O.Print();
    printf("%.6f\n\n", r);
  }
} A, B, C;
double sqr(double x) { return x * x; }
double dist(Point A, Point B) { return sqrt(sqr(A.x - B.x) + sqr(A.y - B.y)); }
double dist(Circle C, Line L) {
  return sqrt(sqr(dist(C.O, L.p)) - fabs((L.q - L.p) ^ (C.O - L.p)));
}
void check(Point cur) {
  double d = dist(cur, A.O);
  double alpha = asin(A.r / d);
  if (alpha > Maxalpha) Maxalpha = alpha, ans = cur;
}
Point Intersection(Line A, Line B, int cc = 1) {
  Point ret = A.p;
  double u = (A.p - B.p) * (B.p - B.q);
  double v =
      (A.p.x - A.q.x) * (B.p.y - B.q.y) - (A.p.y - A.q.y) * (B.p.x - B.q.x);
  if (Zero(v)) return (Point){1e99, 1e99};
  u /= v;
  ret.x += (A.q.x - A.p.x) * u;
  ret.y += (A.q.y - A.p.y) * u;
  if (cc) check(ret);
  return ret;
}
int Intersection(Circle C, Line L, Point &A, Point &B, int cc = 1) {
  Point p = C.O;
  p.x += L.p.y - L.q.y;
  p.y += L.q.x - L.p.x;
  p = Intersection((Line){p, C.O}, L, 0);
  double tmp = sqr(C.r) - sqr(p.x - C.O.x) - sqr(p.y - C.O.y);
  if (tmp < -EPS) return 0;
  double t = sqrt(max(tmp, 0.)) / dist(L.p, L.q);
  A = p + (L.q - L.p) * t;
  B = p - (L.q - L.p) * t;
  if (cc) check(A), check(B);
  return 1;
}
int Intersection(Circle C1, Circle C2, Point &A, Point &B, int cc = 1) {
  if (C1.r < C2.r) swap(C1, C2);
  double d = dist(C1.O, C2.O);
  if (Zero(C1.r + C2.r - d))
    return A = B = C1.O + (C2.O - C1.O) * (C1.r / (C1.r + C2.r)), 1;
  if (Zero(C1.r - C2.r - d))
    return A = B = C2.O + (C2.O - C1.O) * (C2.r / (C1.r - C2.r)), 1;
  if (C1.r + C2.r < d || d < C1.r - C2.r) return 0;
  double t = (1 + (sqr(C1.r) - sqr(C2.r)) / sqr(dist(C1.O, C2.O))) / 2.;
  Point u = C1.O + (C2.O - C1.O) * t;
  Point v = (Point){u.x + C1.O.y - C2.O.y, u.y - C1.O.x + C2.O.x};
  Intersection(C1, (Line){u, v}, A, B, 0);
  check(A);
  check(B);
  return 1;
}
Line solve1(Circle A, Circle B) {
  double tmp1 = 2 * (sqr(B.r) * A.O.x - sqr(A.r) * B.O.x);
  double tmp2 = 2 * (sqr(B.r) * A.O.y - sqr(A.r) * B.O.y);
  double con = sqr(A.r) * (sqr(B.O.x) + sqr(B.O.y)) -
               sqr(B.r) * (sqr(A.O.x) + sqr(A.O.y));
  if (Zero(tmp1))
    return (Line){(Point){0, -con / tmp2}, (Point){1, -con / tmp2}};
  if (Zero(tmp2))
    return (Line){(Point){-con / tmp1, 0}, (Point){-con / tmp1, 1}};
  Point u, v;
  u.x = 1;
  u.y = (-con - u.x * tmp1) / tmp2;
  v.x = 2;
  v.y = (-con - v.x * tmp1) / tmp2;
  return (Line){u, v};
}
Circle solve2(Circle A, Circle B) {
  double tmp1 = 2 * (sqr(B.r) * A.O.x - sqr(A.r) * B.O.x);
  double tmp2 = 2 * (sqr(B.r) * A.O.y - sqr(A.r) * B.O.y);
  double con = sqr(A.r) * (sqr(B.O.x) + sqr(B.O.y)) -
               sqr(B.r) * (sqr(A.O.x) + sqr(A.O.y));
  double xi = sqr(A.r) - sqr(B.r);
  tmp1 /= xi;
  tmp2 /= xi;
  con /= xi;
  Circle ret;
  ret.O = (Point){-tmp1 / 2., -tmp2 / 2.};
  ret.r = sqr(tmp1 / 2.) + sqr(tmp2 / 2.) - con;
  ret.r = sqrt(ret.r);
  return ret;
}
int main() {
  A.read();
  B.read();
  C.read();
  Line BC1;
  Circle BC2;
  Point t1, t2;
  if (Zero(A.r - B.r)) {
    Line AB = solve1(A, B);
    if (Zero(B.r - C.r))
      Intersection(AB, solve1(B, C));
    else
      Intersection(solve2(B, C), AB, t1, t2);
  } else {
    Circle AB = solve2(A, B);
    if (Zero(B.r - C.r))
      Intersection(AB, solve1(B, C), t1, t2);
    else
      Intersection(solve2(B, C), AB, t1, t2);
  }
  if (Maxalpha > -0.5) ans.Print();
}